Math Is Fun Forum

ant

Member

Registered: 2006-06-07

Posts: 9

absolute value inequality

Hi im having a bit of trouble with an absolute value inequality...

l x² - 1 l ≤ 4

i know that sounds dumb, but because of the squared? i'm rather confused...the answer reckons its    -√5 ≥ x ≥ √5
but i just don't agree. when you solve don't you get -3 on the left?
and then what do you do about the fact that you can't take the root of a negative number? i'm not sure what to do after

-4 ≤ (x² - 1) ≤ 4

-3 ≤ x² ≤ 5

thanks a bunch!!!!

Last edited by ant (2006-10-06 01:34:26)

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If you're not part of the solution, you're part of the precipitate.

ant

Member

Registered: 2006-06-07

Posts: 9

Re: absolute value inequality

ok i might have just answered my own question, but could someone please check?

can you do something by saying

l x² - 1 l = 4

x² - 1 = +/- 4

then

x² - 1 = - 4
x² = -3                 so NO SOLUTION

and on the other side:

x² - 1 = +4
x² = 5

x = =/-√5

and then cos it wasnt equal to, it was less than or equal to, you then say

-√5 ≥ x ≥ √5

or you could graph it to check this.

does this work?

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If you're not part of the solution, you're part of the precipitate.

fgarb

Member

Registered: 2006-03-03

Posts: 89

Re: absolute value inequality

Looks right to me. When you say x^2=-3, you're not including the inequality part - what it's really saying is x^2>=-3, which is of course always satisfied, so you can ignore it. Congrats on figuring it out yourself!

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: absolute value inequality

-3 ≥ x² ≥ 5 ?

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: absolute value inequality

You can actually ignore the modulus sign on this occasion. x²-1 will never be less than -4, so the ≥ -4 restriction doesn't matter.

And if you just solve x²-1≤4, then you get x²≤5 and so -√5 ≤x≤√5.

Edit: Off-topic, but I used the same code to write each of those square root signs, and they've come out different to each other. How strange.

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Why did the vector cross the road?
It wanted to be normal.

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: absolute value inequality

I don't remember the rules regarding absolute values, inequalities, and square roots, but your mistake was going from

There's some rule involving square roots and inequalities.   For example: 

   I know

.   The

is the easy part.   How do I  prove mathematically

?  Anyone?

I'm thinking this through as I go, so bear with me.  Consider a more interesting but very similar problem: 

.   First off, consider only those values of x where

(without absolute value) is a positive number. 
The values of x which satisify this are:

.   For these values of x, the absolute value function has no effect on the equation:

Now, consider those values where

(without the absolute value) results in a negative number. This is the points between

.  When you take the absolute value of a negative number, you're essentially multiplying by -1.   

So your answer must satisfy all of these limitations:

This is all the points between -3 and -1  plus  the points beween 1 and 3.

I didn't figure out your answer and I probably just made things worse but maybe my comments will job someone else's memory.    Bottom line is that the problem is not as simple as it first appears.   You need to manipulate the equations based on the value of x.   I need to stop for now.  Maybe I'll come back later this evening and try to clear up this mess.

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: absolute value inequality

Sorry, but I was a little off there in my first post.  Your step from

was perfectly fine.   So was the next step:

Where I (we?) went wrong was I was looking for the -3 to be the lower bound and 5 to be the upper bound (actually the square roots).   The answer to the above equation is really equal to the intersection of the two equations:

All values of x satisfy the first equation.  For the second, the following set of x's satisfy the equation:

So the answer is the set of x's which meet both criteria which in this case is simply:

A better example to show how the intersection works is to subtract 5 rather than 1, like I did in my first post.

Now you must take the intersection of these two sets of answers.  The first equation limits x to values between -3 and 3.   The second limits x to values greater than 1 or less than -1.  The intersection gives us the following 2 intervals:

Again, sorry about the first post.   Although I don't think any of it was wrong, it just made the waters muddier.  This one probably didn't help much either!

ant

Member

Registered: 2006-06-07

Posts: 9

Re: absolute value inequality

thanks for all the help!
i think im getting it... i get where i went wrong at first (also, in one of the lines, i put all the inequality signs the wrong way round, which wasn't helpful. *remembers that crocodiles always eat the bigger fish*)

so just to reiterate, can you just ignore the part that says  -3 ≤ x² ?

i also graphed it, just to check!

thanks for all the help!

nup. i cant figure out how to post a pic. no probs tho. thanks again

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If you're not part of the solution, you're part of the precipitate.

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: absolute value inequality

so just to reiterate, can you just ignore the part that says  -3 ≤ x² ?

Well, ignore is a pretty strong word.   In your equation, it has no effect on the answer.

Last edited by pi man (2006-10-06 03:48:10)